This exercise considers the problem of finding the traverse displacement
(
) of an elastic beam, which is subject to a body force
(). The potential energy is
where
is a positive constant. It is assumed that the beam has what are called simply supported ends. One consequence of this is that
(0) = 0 and
(1) = 0. The role of the boundary conditions will be discussed in more detail in part (f).
(a) Suppose the grid points are
0
= 0,
1
=
,
2
= 2, ··· ,
n+1
= 1, where
= 1/(
+ 1). Writing
write down the composite trapezoidal approximation for .
(b) Use a centered second-order approximation for
xx
at
1,
2, ··· ,
n
. You can use a first-order approximations for
xx
at
0
and
n+1. Using these with the result from part (a), what is the resulting approximation for ? In doing this, remember that
0
=
n+1
= 0.
(c) What is the resulting matrix equation that must be solved to find the minimum for the approximation for
found in part (b)?
(d) The minimum of
from part (b) can be found using the MATLAB command
where
is the approximation from part (b) and
U
is an
-vector containing a starting guess for (
1,
2, ··· ,
n
)
T
. What would be a good, simple, and nonzero choice for
U, and why is it a good choice? In answering this, it is worth knowing that a cable hanging between two poles is an elastic beam, subject to gravity.
(e) If
= 1 and
=
4, then the exact solution is
On the same axis, plot the exact solution and the numerical solution when using 22 grid points. In doing this, explain how you found the numerical solution (using part (c) or part (d)), and why.
(f) A simply supported beam is required to satisfy the four boundary conditions:
(0) = 0,
(1) = 0,
xx
(0) = 0, and
xx
(1) = 0. The numerical solution was derived without any mention of the last two conditions. The reason is that they are natural boundary conditions, which means that if
minimizes , then it will automatically satisfy
xx
(0) = 0 and
xx
(1) = 0. In comparison,
(0) = 0 and
(1) = 0 are essential boundary conditions, which means that we must explicitly require this of our approximation (see part (b)). So, the question is, does your numerical solution satisfy (approximately)
xx
(0) = 0 and
xx
(1) = 0. Using the first-order approximations you used in part (b), from the numerical solution calculate
xx
(0) and
xx
(1), using
= 20, 40, 80, 160. Are your answers consistent with the statement that the minimizer satisfies
xx
(0) = 0 and
xx
(1) = 0?